Question

Line l is the bisector of an angel A and B is any point on l. BP and BQ are perpendicular from B to the arms of angle A .show that(i) triangle ABP~=triangle AQB (ii) BP=BQ or B is equidistant from the arms of angle A

Answer 1

Given: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A.

To Prove: (i) ∆APB ≅ ∆AQB
(ii) BP = BQ

Or

B is equidistant from the arms of ∠A.
Proof:
(i) In ∆APB and ∆AQB,
∠BAP = ∠BAQ
| ∵ l is the bisector of ∠A
AB = AB    | Common
∠BPA = ∠BQA    | Each = 90°
| ∵ BP and BQ are perpendiculars from B to the arms of ∠A
∴ ∆APB ≅ ∆AQB    | AAS Rule

(ii) ∵ ∆APB ≅ ∆AQB
| Proved in (i) above
∴ BP = BQ.   

Answer 2

Given :

• ∠PAB = ∠QAB
• ∠BPA = ∠BQA

To Prove :

• Δ APB ≅ Δ AQB
• BP = BD

Solution :

In Δ APB and Δ AOB :-

$\longmapsto\tt{\angle{PAB}=\angle{QAB}(Given)}$

$\longmapsto\tt{\angle{BPA}=\angle{BQA}(Each\:90\degree)}$

$\longmapsto\tt{AB=AB(Common)}$

So , By ASA Rule ΔAPB ≅ ΔAQB..

Now :

$\longmapsto\tt{BP=BQ(By\:CPCT\:Rule)}$

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ASA Rule :

Two Triangles are congruent if two angles and the included side of One triangle is equal to the two angles and the included side of other Triangle.

CPCT Rule :

CPCT means Corresponding Parts of Congruent Triangles.

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